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Actually a stronger statement can be made:A group G is finite if and only if the number of its subgroups is finiteLet G be a group. If G is finite there is only a finite number of subsets of G, so clearlya finite number of subgroups.Now suppose G is infinite , let'ssuppose one element has infinite order. The this element generates an infinite cyclicgroup which in turn contains infinitely many subgroups.Now suppose all the subgroups have finite order Take some element of G and let it generate a finite group H. Now take another element of G not in H and let it generate a finite group I. Keep doing this by next picking an element of G not H or I. You can continue this way.
By LaGrange's Thm., the order of an element of a group must divide the order of the group. Since 3 is prime, up to isomorphism, the only group of order three is {1,x,x^2} where x^3=1. Note that this is a finite cyclic group. Since all cyclic groups are abelian, because they can be modeled by addition mod an integer, the group of order 3 is abelian.
There are four requirements that need to be satisfied: A. Closure: For any two elements of the group, a and b, the operation a*b is also a member of the group. B. Associativity: For any three members of the group, a*(b*c) = (a*b)*c C. Identity: There exists an element in the group, called the identity and denoted by i, such that a*i = i*a for all a in the group. For real numbers with multiplication, this element is 1. D. Inverse: For any member of the group, a, there exists a member of the group, b, such that a*b = b*a = 1 (the identity). b is called the inverse of a and denoted by a-1.
Let G be the cyclic group generated by x, say. Ten every elt of G is of the form x^a, for some a
The Outsiders is a novel about 2 gangs of kids. One group being known as the greasers and the other group known as socs. compete in rumbles, fights, jumps, to prove who is better.
(1). G is is finite implies o(G) is finite.Let G be a finite group of order n and let e be the identity element in G. Then the elements of G may be written as e, g1, g2, ... gn-1. We prove that the order of each element is finite, thereby proving that G is finite implies that each element in G has finite order. Let gkbe an element in G which does not have a finite order. Since (gk)r is in G for each value of r = 0, 1, 2, ... then we conclude that we may find p, q positive integers such that (gk)p = (gk)q . Without loss of generality we may assume that p> q. Hence(gk)p-q = e. Thus p - q is the order of gk in G and is finite.(2). o(G) is finite implies G is finite.This follows from the definition of order of a group, that is, the order of a group is the number of members which the underlying set contains. In defining the order we are hence assuming that G is finite. Otherwise we cannot speak about quantity.Hope that this helps.
Actually a stronger statement can be made:A group G is finite if and only if the number of its subgroups is finiteLet G be a group. If G is finite there is only a finite number of subsets of G, so clearlya finite number of subgroups.Now suppose G is infinite , let'ssuppose one element has infinite order. The this element generates an infinite cyclicgroup which in turn contains infinitely many subgroups.Now suppose all the subgroups have finite order Take some element of G and let it generate a finite group H. Now take another element of G not in H and let it generate a finite group I. Keep doing this by next picking an element of G not H or I. You can continue this way.
False. G may be a finite group without sub-groups.
By LaGrange's Thm., the order of an element of a group must divide the order of the group. Since 3 is prime, up to isomorphism, the only group of order three is {1,x,x^2} where x^3=1. Note that this is a finite cyclic group. Since all cyclic groups are abelian, because they can be modeled by addition mod an integer, the group of order 3 is abelian.
The order of a group is the same as its cardinality - i.e. the number of elements the set contains. The order of a particular element is the order of the (cyclic) group generated by that element - i.e. the order of the group {...a-4, a-3, a-2, a-1, e, a, a2, a3, a4...}. If these powers do not go on forever, it will have a finite order; otherwise the order will be infinite.
Normally, a cyclic group is defined as a set of numbers generated by repeated use of an operator on a single element which is called the generator and is denoted by g.If the operation is multiplicative then the elements are g0, g1, g2, ...Such a group may be finite or infinite. If for some integer k, gk = g0 then the cyclic group is finite, of order k. If there is no such k, then it is infinite - and is isomorphic to Z(integers) with the operation being addition.
yes form cayleys theorem . every group is isomorphic to groups of permutation and finite groups are not an exception.
The order of a group is the same as its cardinality - i.e. the number of elements the set contains. The order of a particular element is the order of the (cyclic) group generated by that element - i.e. the order of the group {...a-4, a-3, a-2, a-1, e, a, a2, a3, a4...}. If these powers do not go on forever, it will have a finite order; otherwise the order will be infinite.
In group theory, an alternating group is a group of even permutations of a finite set.
Wolfgang Hamernik has written: 'Group algebras of finite groups' -- subject(s): Finite groups, Group algebras
You would use trend of the periodic table to pinpoint the group or period of the substance. Then you would use group/period characteristics. If you can not find an element that matches, you will have to repeat the process which you created the element(if you created it). If you found the element then search around for more and team-up with a renowned chemist and prove it. ***It could take a few years to a coule decades to prove it***
The order of a group is the same as its cardinality - i.e. the number of elements the set contains. The order of a particular element is the order of the (cyclic) group generated by that element - i.e. the order of the group {...a-4, a-3, a-2, a-1, e, a, a2, a3, a4...}. If these powers do not go on forever, it will have a finite order; otherwise the order will be infinite.